Study the circuit diagram in the figure below and answer the questions that follow.

The 4Ω and 3Ω resistors are in parallel.

$$R_p = \frac{R_1 \times R_2}{R_1 + R_2}$$ $$R_p = \frac{4\Omega \times 3\Omega}{4\Omega + 3\Omega}$$ $$R_p = \frac{12\Omega^2}{7\Omega}$$ $$R_p = \frac{12}{7}\Omega$$

The 2Ω resistor is in series with the parallel combination $R_p$.

$$R_{total} = R_{series} + R_p$$ $$R_{total} = 2\Omega + \frac{12}{7}\Omega$$ $$R_{total} = \frac{14}{7}\Omega + \frac{12}{7}\Omega$$ $$R_{total} = \frac{26}{7}\Omega$$ $$\boxed{R_{total} = \frac{26}{7}\Omega \approx 3.71\Omega}$$

II. The current through the 4Ω resistor

Using Ohm's Law, $V = I R_{total}$.

$$I_{total} = \frac{V_{source}}{R_{total}}$$ $$I_{total} = \frac{12V}{\frac{26}{7}\Omega}$$ $$I_{total} = \frac{12 \times 7}{26} A$$ $$I_{total} = \frac{84}{26} A$$ $$I_{total} = \frac{42}{13} A$$

The total current $I_{total}$ flows through the 2Ω resistor. The voltage drop across the 2Ω resistor ($V_{2\Omega}$) is:

$$V_{2\Omega} = I_{total} \times 2\Omega$$ $$V_{2\Omega} = \frac{42}{13}A \times 2\Omega$$ $$V_{2\Omega} = \frac{84}{13}V$$

The voltage across the parallel combination ($V_p$) is the source voltage minus the voltage drop across the 2Ω resistor:

$$V_p = V_{source} - V_{2\Omega}$$ $$V_p = 12V - \frac{84}{13}V$$ $$V_p = \frac{156}{13}V - \frac{84}{13}V$$ $$V_p = \frac{72}{13}V$$

The voltage across the 4Ω resistor is $V_p$.

$$I_{4\Omega} = \frac{V_p}{4\Omega}$$ $$I_{4\Omega} = \frac{\frac{72}{13}V}{4\Omega}$$ $$I_{4\Omega} = \frac{72}{13 \times 4} A$$ $$I_{4\Omega} = \frac{18}{13} A$$ $$\boxed{I_{4\Omega} = \frac{18}{13}A \approx 1.38A}$$